Sodium acetate is the salt of a strong base and a weak acid so we would expect it to be slightly alkali due to salt hydrolysis:
sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^-)
For which:
sf(K_b=([CH_3COOH][OH^-])/([CH_3COO^-]))
Before we set up an ICE table we need to find its initial concentration.
The number of moles is given by:
sf(n_(CH_3COONa)=m/M_r=4.0/82.034=0.0487)
We get the initial concentration using sf(c=n/v):
sf([CH_3COONa]=0.0487/0.500=0.09752color(white)(x)"mol/l")
Now set up the ICE table based on sf("mol/l"):
sf(" "CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^-)
sf(I" "0.09752" "0" "0)
sf(C" "-x" "+x" "+x)
sf(E" "(0.09752-x)" "x" "x)
:.sf(K_b=(x^2)/((0.09752)-x))
Since we are given sf(pK_a) we use this expression:
sf(pK_a+pK_b=14)
:.sf(pK_b=14-4.74=9.26)
This means sf(K_b) is a very small number, so small that we can make the approximation that sf((0.09752-x)rArr0.09752)
So we can write:
sf(K_b=x^2/0.09752)
:.sf(x^2=K_bxx0.09752)
sf(x=sqrt(K_bxx0.09752)
Since sf(x=[OH^-]) we can take negative logs of both sides to get:
sf(pOH=1/2[pK_b-log(0.09752)])
sf(pOH=1/2[9.26-(-1.01090)])
sf(pOH=5.1354)
Since sf(pH+pOH=14) we get:
sf(pH=14-pOH=14-5.1354=8.86)
As expected, the pH is slightly alkaline.
